Question

question...

Answer 1

\[4^x=(2^x)^2\]
is that right?

Answer 2

\[(x^{m})^n=x^{mn}\]
So \[(4^x)^2=4^{2x}\]

Answer 3

Yeah but,
\[4^x=(2^2)^x=2^{2x}=(2^x)^2\]
Would that not be correct?
I find it interesting how you can manipulate the powers like that

Answer 4

Thats actually pretty cooolllll. Why didnt i ever notice that
\[4^x=(2^2)^x=2^{2x}=2^{x+x}=2^x*2^x=(2^x)^2\]

Answer 5

Ehhh we kinda did ... I think we have been doing it without realizing tbh

Answer 6

Hmm, here's a question using that concept
can you find the derivative?
\[y=\tan^-1(\frac{ 2^{x+1} }{ 1-4^x })\]
hint:use substitution :)

Answer 7

Interesting question, I solved it already though

Answer 8

@Nishant_Garg , similar to your other Q's :)

2^x = tan(alpha)

Answer 9

yeah

Answer 10

@wahahaha , can you solve it?
have you studied trig?

Answer 11

My apologies if you have not reached this far @wahahaha

Answer 12

Ok like i most probably came across this in calc 1 but have not done this stuff in like 3-4 years
I just wanna watch u guys solve it and try to refresh my memory

Answer 13

Do you use trig substitution?

Answer 14

or just regular substitution?

Answer 15

haha oh ok , yeah this would be calc 1

you're just studying finance then?

Answer 16

trig subs

Answer 17

Idk what the system is over there, but im in grade 12

Answer 18

ohhhh -.-
In grade 12 the most we got to was differentiating

Ya I am just doing the CFA exam but before that I was studying math which as youc an see im pretty rusty at

Answer 19

math is great but finance gives better opportunities, i wish i had done more in business and finance
good luck with your exam

Answer 20

in grade 12 we got over here
relations and functions
inverse trig
matrices
determinants
differential calculus(limits, continuity, rolls thm and MVT, differentials and approximations, increasing and decreasing functions, maxima and minima, 2nd order derivaives,tangents and normals)
integral calculus(indefinite integrals, limit of sum, definite integrals,area under curves)
differential equations(homogenous equations, first order equations with integrating factor)
vectors and 3d geometry
linear programming
probability(conditional, bayes thm, mean and variance of a random variable,binomial distribution)

Answer 21

worst is 3d so many formulae :(

Answer 22

Ya I felt like math was limiting and I cldnt get a job based on that and so I decided to turn to finance hoping that I can find a job in that field. It is hard to break into the market though. SO many ppl with MBAs now a days so there is alot of competition and you gotta somehow try to stand out

Answer 23

@Nishant_Garg , thats pretty in depth.
I would say only the advanced 12th grade students here have those courses

Answer 24

@Nishant_Garg can you show me how u wld solve this?

Answer 25

@dumbow that's the course for mathematics for grade 12, any student that takes science (computer or medical) or any stream that has maths, it's the same syllabus

Answer 26

sure

Answer 27

\[y=\tan^-1(\frac{ 2.2^x }{ 1-(2^x)^2 })\]
\[2^x=\tan \alpha\]
\[\alpha=\tan^-1(2^x)\]\[y=\tan^-1(\frac{ 2\tan \alpha }{ 1-\tan^2 \alpha })\]\[y=\tan^-1(\tan(2\alpha))\]\[y=2\alpha\]\[y=2\tan^-1(2^x)\]\[\frac{ dy }{ dx }=2 \frac{ 1 }{ 1+(2^x)^2 } \frac{ d(2^x) }{ dx }\]\[\frac{ dy }{ dx }=2\frac{ 1 }{ 1+4^x }2^xlog2\]\[\frac{ dy }{ dx }=\frac{ 2^{x+1}\log2 }{ 1+4^x }\]

Answer 28

@wahahaha here u go